AP Physics 2 Practice Test 6 – 60 Questions with Answers

Test yourself with 60 original AP Physics 2 practice questions covering all 7 units of the course: thermodynamics, electric force, field, and potential, electric circuits, magnetism and electromagnetism, geometric optics, waves, sound, and physical optics, and modern physics. Every answer includes the full working. All questions are algebra-based (no calculus). Take g = 9.8 m/s2 unless stated otherwise.

Unit 9: Thermodynamics

Q1. A sample of air trapped in a bicycle pump is initially at 120 kPa, occupies 6 L and is at 300 K. It is then compressed or expanded to 3 L and its temperature changes to 300 K. What is the final pressure of the gas?

  • (A) 480 kPa
  • (B) 120 kPa
  • (C) 360 kPa
  • (D) 240 kPa
Show Answer

Answer: D – P1V1 ÷ T1 = P2V2 ÷ T2, so P2 = P1 × (V1 ÷ V2) × (T2 ÷ T1) = 120 × (6 ÷ 3) × (300 ÷ 300) = 240 kPa.

Q2. 500 J of heat is added to a gas in a piston while the gas does 200 J of work on its surroundings. What is the change in internal energy of the gas?

  • (A) 300 J
  • (B) 700 J
  • (C) -300 J
  • (D) 500 J
Show Answer

Answer: A – First law: ΔU = Q – Wby = 500 – (200) = 300 J.

Q3. A gas in a piston expands at a constant pressure of 200 kPa from 3 L to 7 L. How much work is done by the gas?

  • (A) 1600 J
  • (B) 400 J
  • (C) 2000 J
  • (D) 800 J
Show Answer

Answer: D – W = PΔV = 200 kPa × (7 – 3) L = 200 × 4 = 800 J, since 1 kPa × 1 L = 1 J.

Q4. How much heat must be supplied to a 3 kg aluminium block of specific heat capacity 900 J/kg·K to raise its temperature by 25 K? (Assume no heat is lost to the surroundings.)

  • (A) 135,000 J
  • (B) 33,750 J
  • (C) 67,500 J
  • (D) 2,700 J
Show Answer

Answer: C – Q = mcΔT = 3 × 900 × 25 = 67,500 J.

Q5. A heat engine does 300 J of useful work in each cycle while absorbing 1200 J of heat from a hot reservoir. What is the efficiency of the engine?

  • (A) 75%
  • (B) 25%
  • (C) 50%
  • (D) 12.5%
Show Answer

Answer: B – Efficiency = W ÷ Qh = 300 ÷ 1200 = 0.25 = 25%.

Q6. An ideal Carnot engine operates between reservoirs at 600 K and 300 K. What is its maximum possible efficiency?

  • (A) 25%
  • (B) None of the other options
  • (C) It cannot be determined from the data given
  • (D) 50%
Show Answer

Answer: D – Carnot efficiency = 1 – Tc ÷ Th = 1 – 300 ÷ 600 = 1 – 0.5 = 0.5 = 50%.

Q7. The root-mean-square speed of the molecules in a gas sample is 400 m/s. If the absolute temperature of the gas is multiplied by 4 while the gas stays the same, what is the new root-mean-square speed?

  • (A) 1600 m/s
  • (B) 400 m/s
  • (C) 800 m/s
  • (D) None of the other options
Show Answer

Answer: C – vrms is proportional to √T, so the speed is multiplied by √4 = 2. New vrms = 400 × 2 = 800 m/s.

Q8. A block of ice melts in a warm room. What happens to the entropy of the ice as it melts?

  • (A) It decreases
  • (B) It stays the same
  • (C) It increases
  • (D) It becomes zero
Show Answer

Answer: C – Entropy measures disorder. Melting spreads the molecules over more arrangements, so entropy rises.

Q9. What is the internal energy of 2 mol of a monatomic ideal gas at 300 K? (Use R = 8.31 J/mol·K.)

  • (A) 7479 J
  • (B) 14958 J
  • (C) 3739.5 J
  • (D) 4986 J
Show Answer

Answer: A – For a monatomic ideal gas, U = (3 ÷ 2)nRT = 1.5 × 2 × 8.31 × 300 = 7479 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two point charges of 2 μC and 3 μC are placed 0.5 m apart. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 0.43 N
  • (B) 0.11 N
  • (C) 0.86 N
  • (D) 0.22 N
Show Answer

Answer: D – F = kq1q2 ÷ r2 = 9.0 × 109 × (2 × 10-6) × (3 × 10-6) ÷ 0.52 = 0.22 N.

Q11. What is the magnitude of the electric field at a point 0.2 m from a point charge of 4 μC? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 1,800,000 N/C
  • (B) 450,000 N/C
  • (C) 3,600,000 N/C
  • (D) 900,000 N/C
Show Answer

Answer: D – E = kQ ÷ r2 = 9.0 × 109 × (4 × 10-6) ÷ 0.22 = 900,000 N/C.

Q12. What is the electric potential at a point 0.5 m from a point charge of 5 μC, taking potential at infinity to be zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 90,000 V
  • (B) 180,000 V
  • (C) 45,000 V
  • (D) 22,500 V
Show Answer

Answer: A – V = kQ ÷ r = 9.0 × 109 × (5 × 10-6) ÷ 0.5 = 90,000 V.

Q13. A charge of 3 μC is placed in a uniform electric field of 2000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.01 N
  • (B) 0.06 N
  • (C) 0 N
  • (D) None of the other options
Show Answer

Answer: A – F = qE = (3 × 10-6) × 2000 = 0.01 N.

Q14. Two parallel metal plates are 0.06 m apart and have a potential difference of 120 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 4,000 V/m
  • (B) 1,000 V/m
  • (C) 7 V/m
  • (D) 2,000 V/m
Show Answer

Answer: D – E = V ÷ d = 120 ÷ 0.06 = 2,000 V/m.

Q15. How much work is done by the electric field when a charge of 2 μC moves through a potential drop of 500 V?

  • (A) 0.01 J
  • (B) 0 J
  • (C) None of the other options
  • (D) It cannot be determined from the data given
Show Answer

Answer: B – W = qΔV = (2 × 10-6) × 500 = 0 J.

Q16. A proton is placed in a uniform electric field pointing east. In which direction is the electric force on the proton?

  • (A) West
  • (B) North
  • (C) East
  • (D) South
Show Answer

Answer: C – The force on a positive charge is in the direction of the field, while the force on a negative charge is opposite to the field. The field itself points away from positive charges and toward negative charges.

Q17. The distance between two point charges is doubled, with both charges unchanged. How does the electrostatic force between them change?

  • (A) It becomes 1/2 of the original force
  • (B) It becomes 1/4 of the original force
  • (C) It becomes 2 times the original force
  • (D) It becomes 4 times the original force
Show Answer

Answer: B – F is proportional to 1 ÷ r2, so doubling r divides F by 4.

Q18. A proton starting from rest is accelerated through a potential difference of 150 V. What kinetic energy does it gain, in electronvolts?

  • (A) 300 eV
  • (B) 150 eV
  • (C) 75 eV
  • (D) 600 eV
Show Answer

Answer: B – Kinetic energy gained = qV. A proton or electron carries one elementary charge, so KE = 150 eV.

Unit 11: Electric Circuits

Q19. A resistor carries a current of 3 A when a potential difference of 24 V is applied across it. What is its resistance?

  • (A) 72 Ω
  • (B) 0.125 Ω
  • (C) 8 Ω
  • (D) 27 Ω
Show Answer

Answer: C – R = V ÷ I = 24 ÷ 3 = 8 Ω.

Q20. Three resistors of 4 Ω, 6 Ω and 10 Ω are connected in series. What is their equivalent resistance?

  • (A) 10 Ω
  • (B) 20 Ω
  • (C) 12 Ω
  • (D) 30 Ω
Show Answer

Answer: B – For series resistors, Req = R1 + R2 + R3 = 4 + 6 + 10 = 20 Ω.

Q21. Two resistors of 6 Ω and 3 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 2 Ω
  • (B) 9 Ω
  • (C) 4 Ω
  • (D) 2.25 Ω
Show Answer

Answer: A – For two parallel resistors, Req = (R1R2) ÷ (R1 + R2) = (6 × 3) ÷ (6 + 3) = 18 ÷ 9 = 2 Ω.

Q22. A lamp operates at 12 V and draws a current of 2 A. What power does it dissipate?

  • (A) 6 W
  • (B) 48 W
  • (C) 24 W
  • (D) 14 W
Show Answer

Answer: C – P = VI = 12 × 2 = 24 W.

Q23. A battery of 30 V is connected across two resistors of 6 Ω and 12 Ω in series. What is the potential difference across the 6 Ω resistor?

  • (A) 20 V
  • (B) 5 V
  • (C) None of the other options
  • (D) 10 V
Show Answer

Answer: D – The same current flows through both resistors, so V1 = V × R1 ÷ (R1 + R2) = 30 × 6 ÷ 18 = 10 V.

Q24. A 12 V battery is connected across two resistors in parallel. One branch has resistance 6 Ω. What current flows in that branch?

  • (A) 2 A
  • (B) 4 A
  • (C) 1 A
  • (D) 18 A
Show Answer

Answer: A – Each parallel branch has the full battery voltage, so I = V ÷ R = 12 ÷ 6 = 2 A.

Q25. A capacitor of capacitance 20 μF is charged to 12 V. How much charge is stored on it?

  • (A) 32 μC
  • (B) 1.67 μC
  • (C) 480 μC
  • (D) 240 μC
Show Answer

Answer: D – Q = CV = 20 × 12 = 240 μC.

Q26. Two capacitors of 6 μF and 4 μF are connected in parallel. What is their equivalent capacitance?

  • (A) 10 μF
  • (B) 2.4 μF
  • (C) 24 μF
  • (D) 5 μF
Show Answer

Answer: A – Parallel: Ceq = C1 + C2 = 6 + 4 = 10 μF.

Q27. A cell of emf 12 V and internal resistance 1 Ω supplies a current of 2 A. What is the terminal potential difference of the cell?

  • (A) 10 V
  • (B) 12 V
  • (C) 14 V
  • (D) 5 V
Show Answer

Answer: A – Terminal voltage V = emf – Ir = 12 – (2 × 1) = 12 – 2 = 10 V.

Unit 12: Magnetism and Electromagnetism

Q28. A particle with charge 4 μC moves at 3,000 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the particle?

  • (A) 0.06 N
  • (B) 0 N
  • (C) 0.01 N
  • (D) None of the other options
Show Answer

Answer: C – F = qvB = (4 × 10-6) × 3,000 × 0.5 = 0.01 N.

Q29. A straight wire of length 0.6 m carries a current of 3 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the wire?

  • (A) 1.8 N
  • (B) 0.9 N
  • (C) 0.45 N
  • (D) 9 N
Show Answer

Answer: B – F = ILB = 3 × 0.6 × 0.5 = 0.9 N.

Q30. What is the magnitude of the magnetic field at a distance of 0.12 m from a long straight wire carrying a current of 6 A?

  • (A) 10 μT
  • (B) 20 μT
  • (C) 5 μT
  • (D) 100 μT
Show Answer

Answer: A – B = μ0I ÷ (2πr) = (2 × 10-7 × 6) ÷ 0.12 = 10 × 10-6 T = 10 μT.


Q31. A long solenoid has 800 turns per metre and carries a current of 2.5 A. What is the magnitude of the magnetic field inside the solenoid? (Use μ0 = 4π × 10-7 T·m/A and π = 3.14.)

  • (A) 5.02 mT
  • (B) 1.26 mT
  • (C) 2.51 mT
  • (D) 25.12 mT
Show Answer

Answer: C – B = μ0nI = 4π × 10-7 × 800 × 2.5 = 2.51 × 10-3 T = 2.51 mT.

Q32. A metal rod of length 0.4 m moves at 3 m/s perpendicular to a uniform magnetic field of 0.5 T, with the rod perpendicular to both the field and its motion. What emf is induced between the ends of the rod?

  • (A) 0.6 V
  • (B) 1.2 V
  • (C) 0.3 V
  • (D) 6 V
Show Answer

Answer: A – Motional emf = BLv = 0.5 × 0.4 × 3 = 0.6 V.

Q33. The magnetic flux through a coil of 100 turns changes by 0.02 Wb in 0.5 s. What is the magnitude of the average induced emf in the coil?

  • (A) 8 V
  • (B) 2 V
  • (C) 4 V
  • (D) 0.4 V
Show Answer

Answer: C – Faraday’s law: emf = N × (Δflux ÷ time) = 100 × (0.02 ÷ 0.5) = 100 × 0.04 = 4 V.

Q34. A flat loop of area 0.5 m2 is placed in a uniform magnetic field of 0.4 T. The field makes an angle of 0° with the normal to the loop. What is the magnetic flux through the loop?

  • (A) 0.4 Wb
  • (B) 0.2 Wb
  • (C) 0.1 Wb
  • (D) None of the other options
Show Answer

Answer: B – Flux = BA cos θ = 0.4 × 0.5 × cos 0° = 0.2 × 1 = 0.2 Wb.

Q35. A bar magnet with its north pole facing a closed loop is pushed toward the loop. What magnetic pole does the side of the loop facing the magnet become, according to Lenz’s law?

  • (A) A north pole, so that it repels the approaching magnet
  • (B) A south pole, so that it attracts the approaching magnet
  • (C) It does not become a pole at all
  • (D) A north pole, so that it attracts the approaching magnet
Show Answer

Answer: A – Lenz’s law: an induced current always opposes the change that produced it. An approaching pole is repelled, a receding pole is attracted, and a falling magnet is slowed by the opposing induced currents.

Unit 13: Geometric Optics

Q36. An object is placed 30 cm in front of a concave shaving mirror of focal length 10 cm. How far from the mirror is the image formed?

  • (A) 30 cm
  • (B) 15 cm
  • (C) 7.5 cm
  • (D) 20 cm
Show Answer

Answer: B – Mirror equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 10 – 1 ÷ 30 = 0.07 cm-1, giving di = 15 cm.

Q37. An object is placed 36 cm in front of a converging lens of focal length 12 cm. How far from the lens is the image formed?

  • (A) 36 cm
  • (B) 9 cm
  • (C) 18 cm
  • (D) 48 cm
Show Answer

Answer: C – Lens equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 12 – 1 ÷ 36, giving di = 18 cm.

Q38. An object is 20 cm from a mirror and its image is formed 60 cm from the mirror. What is the lateral magnification of the image?

  • (A) 3
  • (B) -3
  • (C) -6
  • (D) -1.5
Show Answer

Answer: B – Magnification m = -di ÷ do = -60 ÷ 20 = -3. The negative sign shows the image is inverted.

Q39. A ray of light travels from a medium of refractive index 1 into a medium of refractive index 1.5 at an angle of incidence of 30°. What is the angle of refraction, to the nearest 0.1°?

  • (A) 29.5°
  • (B) 19.5°
  • (C) 9.5°
  • (D) 70.5°
Show Answer

Answer: B – Snell law: n1 sin θ1 = n2 sin θ2, so sin θ2 = (1 × sin 30°) ÷ 1.5 = 0.33. Hence θ2 = 19.5°.

Q40. What is the critical angle for light travelling from glass of refractive index 1.5 into air, to the nearest 0.1°?

  • (A) 41.8°
  • (B) 51.8°
  • (C) 31.8°
  • (D) 48.2°
Show Answer

Answer: A – At the critical angle, sin θc = n2 ÷ n1 = 1 ÷ 1.5 = 0.67, so θc = 41.8°.

Q41. Light travels at 2 × 108 m/s in a transparent material. What is the refractive index of the material? (Take c = 3.00 × 108 m/s.)

  • (A) 3
  • (B) 0.75
  • (C) 1.5
  • (D) 1
Show Answer

Answer: C – n = c ÷ v = (3.00 × 108) ÷ (2 × 108) = 1.5.

Q42. Light of wavelength 600 nm in air enters a medium of refractive index 1.5. What is its wavelength in the medium?

  • (A) 600 nm
  • (B) 800 nm
  • (C) 400 nm
  • (D) 200 nm
Show Answer

Answer: C – Wavelength in a medium: λ = λ0 ÷ n = 600 ÷ 1.5 = 400 nm. The frequency stays the same, so the wavelength shortens.

Q43. An object is placed beyond the centre of curvature of a concave mirror. Which description best fits the image formed?

  • (A) Virtual, upright and magnified
  • (B) Real, upright and magnified
  • (C) Virtual, inverted and diminished
  • (D) Real, inverted and diminished
Show Answer

Answer: D – Ray diagrams show: a concave mirror gives a real, inverted, diminished image beyond the centre of curvature and a virtual, upright, magnified image inside the focal point; convex mirrors and diverging lenses always give virtual, upright, diminished images; a converging lens with the object beyond 2F gives a real, inverted, diminished image.

Unit 14: Waves, Sound, and Physical Optics

Q44. A wave has frequency 5 Hz and wavelength 20 m. What is its speed?

  • (A) 200 m/s
  • (B) 50 m/s
  • (C) 25 m/s
  • (D) 100 m/s
Show Answer

Answer: D – v = fλ = 5 × 20 = 100 m/s.

Q45. A string of length 2 m is fixed at both ends. Waves travel along it at 240 m/s. What is the frequency of its fundamental?

  • (A) 120 Hz
  • (B) 30 Hz
  • (C) None of the other options
  • (D) 60 Hz
Show Answer

Answer: D – For a string fixed at both ends, fn = nv ÷ (2L) = 1 × 240 ÷ (2 × 2) = 60 Hz.

Q46. An air column of length 0.85 m is open at both ends. Taking the speed of sound as 340 m/s, what is its fundamental frequency?

  • (A) 400 Hz
  • (B) 200 Hz
  • (C) 100 Hz
  • (D) 800 Hz
Show Answer

Answer: B – For a pipe open at both ends, f = v ÷ (2L) = 340 ÷ (2 × 0.85) = 200 Hz.

Q47. Two tuning forks of frequencies 512 Hz and 506 Hz are sounded together. What is the beat frequency heard?

  • (A) 12 Hz
  • (B) 6 Hz
  • (C) 8 Hz
  • (D) 1018 Hz
Show Answer

Answer: B – Beat frequency = |f1 – f2| = |512 – 506| = 6 Hz.

Q48. In a double-slit experiment, light of wavelength 600 nm passes through slits 0.25 mm apart and forms fringes on a screen 2 m away. How far from the central bright fringe is the first bright fringe, in millimetres?

  • (A) 9.6 mm
  • (B) 4.8 mm
  • (C) 2.4 mm
  • (D) 48 mm
Show Answer

Answer: B – Fringe position y = mλL ÷ d = 1 × (600 × 10-9) × 2 ÷ (0.25 × 10-3) = 4.8 × 10-3 m = 4.8 mm.

Q49. A sound source emitting 500 Hz moves at 34 m/s toward a stationary observer. Taking the speed of sound as 340 m/s, what frequency does the observer hear, to the nearest hertz?

  • (A) 596 Hz
  • (B) 516 Hz
  • (C) 556 Hz
  • (D) 500 Hz
Show Answer

Answer: C – Doppler shift for a moving source: f’ = f × v ÷ (v – vs) = 500 × 340 ÷ 306 = 556 Hz to the nearest hertz.

Q50. A listener moves to twice the original distance from a point source of sound, with no other change. How does the sound intensity at the listener change?

  • (A) It becomes 1/2 of the original intensity
  • (B) It becomes 2 times the original intensity
  • (C) It becomes 1/4 of the original intensity
  • (D) It stays the same
Show Answer

Answer: C – Intensity follows an inverse-square law, I proportional to 1 ÷ r2, so doubling r divides I by 4.

Q51. A diffraction grating has 500 lines per millimetre. Light of wavelength 600 nm is incident normally on it. What is the angle of the first-order maximum, to the nearest 0.1°?

  • (A) 27.5°
  • (B) 17.5°
  • (C) 7.5°
  • (D) 72.5°
Show Answer

Answer: B – Grating spacing d = 1 ÷ 500 mm = 2000 nm. For the first order, d sin θ = λ, so sin θ = 600 ÷ 2000 = 0.3, giving θ = 17.5°.

Unit 15: Modern Physics

Q52. What is the energy, in electronvolts, of a photon of wavelength 620 nm? (Use hc = 1240 eV·nm.)

  • (A) 2 eV
  • (B) 4 eV
  • (C) 1 eV
  • (D) 3 eV
Show Answer

Answer: A – E = hc ÷ λ = 1240 ÷ 620 = 2 eV.

Q53. Light of photon energy 4 eV falls on a metal surface whose work function is 2.3 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 1.7 eV
  • (B) 6.3 eV
  • (C) 4 eV
  • (D) 3.4 eV
Show Answer

Answer: A – Einstein photoelectric equation: Kmax = hf – work function = 4 – 2.3 = 1.7 eV.

Q54. A metal has a work function of 2 eV. What is its threshold frequency, to two decimal places in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s and 1 eV = 1.60 × 10-19 J.)

  • (A) 9.65 × 1014 Hz
  • (B) 2.41 × 1014 Hz
  • (C) 7.24 × 1014 Hz
  • (D) 4.83 × 1014 Hz
Show Answer

Answer: D – Threshold: hf0 = work function = 2 × 1.60 × 10-19 J = 3.2 × 10-19 J. So f0 = (3.2 × 10-19) ÷ (6.63 × 10-34) = 4.83 × 1014 Hz.

Q55. An electron moves at 1 × 106 m/s. What is its de Broglie wavelength, in nanometres? (Use h = 6.63 × 10-34 J·s and electron mass 9.11 × 10-31 kg.)

  • (A) 1.46 nm
  • (B) 0.36 nm
  • (C) 7.28 nm
  • (D) 0.73 nm
Show Answer

Answer: D – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 1 × 106) = 0.73 × 10-9 m = 0.73 nm.

Q56. A hydrogen atom makes a transition from energy level n = 3 to n = 2. How much energy is released as a photon, in electronvolts? (Use En = -13.6 ÷ n2 eV.)

  • (A) 3.78 eV
  • (B) 0.94 eV
  • (C) 11.71 eV
  • (D) 1.89 eV
Show Answer

Answer: D – Energy released = E3 – E2 in magnitude = 13.6 × (1 ÷ 22 – 1 ÷ 32) = 13.6 × 0.14 = 1.89 eV.

Q57. A radioactive sample initially contains 80 g of an isotope. After 3 half-lives, how much of the isotope remains?

  • (A) 20 g
  • (B) 10 g
  • (C) 5 g
  • (D) 70 g
Show Answer

Answer: B – After each half-life half remains, so after 3 half-lives the fraction left is (1 ÷ 2)3 = 1 ÷ 8. Remaining mass = 80 ÷ 8 = 10 g.

Q58. In a nuclear reaction the total mass decreases by 0.02 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 18.63 MeV
  • (B) 37.26 MeV
  • (C) 9.31 MeV
  • (D) 186.3 MeV
Show Answer

Answer: A – E = Δmc2 = 0.02 × 931.5 = 18.63 MeV.

Q59. A clock on a spacecraft moving at 0.6c ticks 8 s as measured in the spacecraft frame. How much time passes on Earth while the spacecraft clock ticks 8 s?

  • (A) 8 s
  • (B) 20 s
  • (C) 10 s
  • (D) 16 s
Show Answer

Answer: C – For v = 0.6c, the Lorentz factor is 1.25. Dilated time = 1.25 × 8 = 10 s.

Q60. A spacecraft of proper length 100 m travels past Earth at 0.6c. What length does an observer on Earth measure for the spacecraft?

  • (A) 100 m
  • (B) 160 m
  • (C) 50 m
  • (D) 80 m
Show Answer

Answer: D – Length contraction: L = L0 ÷ Lorentz factor = 100 ÷ 1.25 = 80 m.

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